POJ 2299 逆序对

Crossings

Time Limit: 2 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100463

Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence 
9 1 0 5 4 ,
Ultra-QuickSort produces the output 
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

5

9

1

0

5

4

3

1

2

3

0

Sample Output

6

0

HINT

 

题意

 逆序对求解

题解：

树状数组水

代码

 1 #include <cstdio>

 2 #include <cmath>

 3 #include <cstring>

 4 #include <ctime>

 5 #include <iostream>

 6 #include <algorithm>

 7 #include <set>

 8 #include <vector>

 9 #include <queue>

10 #include <map>

11 #include <stack>

12 #define MOD 1000000007

13 #define maxn 32001

14 using namespace std;

15 typedef __int64 ll;

16 inline ll read()

17 {

18     ll x=0,f=1;

19     char ch=getchar();

20     while(ch<'0'||ch>'9')

21     {

22         if(ch=='-')f=-1;

23         ch=getchar();

24     }

25     while(ch>='0'&&ch<='9')

26     {

27         x=x*10+ch-'0';

28         ch=getchar();

29     }

30     return x*f;

31 }

32 //*******************************************************************

33 

34 struct ss

35 {

36     int v,index;

37 } in[500005];

38 int c[500005];

39 int a[500005];

40 int n;

41 bool cmp(ss s1,ss s2)

42 {

43     return s1.v<s2.v;

44 }

45 int lowbit(int x)

46 {

47     return x&(-x);

48 }

49 int getsum(int x)

50 {

51     int sum=0;

52     while(x>0)

53     {

54         sum+=c[x];

55         x-=lowbit(x);

56     }

57     return sum;

58 }

59 void update(int x,int value)

60 {

61     while(x<=n)

62     {

63         c[x]+=value;

64         x+=lowbit(x);

65     }

66 }

67 int main()

68 {

69 

70     while(scanf("%d",&n)!=EOF)

71     {

72         if(n==0) break;

73         memset(c,0,sizeof(c));

74         for(int i=1; i<=n; i++)

75         {

76             in[i].v=read();

77             in[i].index=i;

78         }

79         sort(in+1,in+n+1,cmp);

80         for(int i=1; i<=n; i++)a[in[i].index]=i; //离散化处理

81         ll ans=0;

82         for(int i=1; i<=n; i++)

83         {

84             update(a[i],1);

85             ans+=i-getsum(a[i]);

86         }

87         cout<<ans<<endl;

88     }

89 

90     return 0;

91 }